Research Article
α-ψ-ϕ-contractive mappings in ordered partial b-metric spaces
Aiman Mukheimera,∗
aDepartment of Mathematics and General Sciences, Prince Sultan University, P.O.Box 66833, Riyadh 11586, Saudi Arabia
Communicated by Imdad Khan
Abstract
In this paper, we introduce the concept ofα-ψ-ϕ-contractive self mapping in complete ordered partialb- metric space, and we study the existence of fixed points for such mappings under some conditions. Presented theorems in this paper extend and generalize the results derived by Mustafa et al., also some examples are given to illustrate the main results. c2014 All rights reserved.
Keywords: b-metric space, fixed point theory, contraction, partial metric space.
2010 MSC: 47H10, 54H25.
1. Introduction and Preliminaries
Fixed point theory is one of the most popular tool in nonlinear analysis. Most of the generalizations for metric fixed point theorems usually start from Banach contraction principle [8]. It is not easy to point out all the generalizations of this principle. In 1989, Bakhtin [7] introduced the concept of a b-metric space as a generalization of metric spaces. In 1993, Czerwik [9, 10] extended many results related to the b−metric spaces. In 1994, Matthews [15] introduced the concept of partial metric space in which the self distance of any point of space may not be zero. In 1996, O’Neill generalized the concept of partial metric space by admitting negative distances. In 2013, Shukla [21] generalized both the concept of b-metric and partial metric spaces by introducing the partialb-metric spaces. For example, many authors recently studied this principle and its generalizations in different types of metric spaces [12, 23, 1, 2, 18, 20, 5]. Close to our interest in this paper some authors studied some fixed point theorems in the so called b−metric space [16, 22, 24]. After then, some authors started to proveα-ψversions of of certain fixed point theorems in different type metric spaces
∗Corresponding author
Email address: [email protected](Aiman Mukheimer) Received 2013-12-23
[3, 11, 4]. Mustafa in [17], gave a generalization of Banach’s contraction principles in a complete ordered partial b-metric space by introducing a generalized (α, ψ)s-weakly contractive mapping. In this paper, we generalize a result of Mustafa in [17], by introducing theα-ψ-ϕ-contractive mapping in a complete ordered partial b-metric space.
Definition 1.1. [6] LetX be a nonempty set and lets≥1 be a given real number. A functiond:X×X → [0,∞) is called ab-metric if for all x, y, z∈X the following conditions are satisfied:
(i) d(x, y) = 0 if and only ifx=y;
(ii)d(x, y) =d(y, x);
(iii) d(x, y)≤s[d(x, z) +d(z, y)].
The pair (X, d) is called ab-metric space. The number s≥1 is called the coefficient of (X, d).
Definition 1.2. [15] LetX be a nonempty set. A function p:X×X →[0,∞) is called a partial metric if for all x, y, z∈X the following conditions are satisfied:
(i) x=y if and only ifp(x, x) =p(x, y) =p(y, y);
(ii)p(x, x)≤p(x, y);
(iii) p(x, y) =p(y, x);
(iv)p(x, y)≤p(x, z) +p(z, y)−p(z, z).
The pair (X, d) is called a partial metric space.
Remark 1.1. It is clear that the partial metric space need not be ab-metric spaces , since in a partial metric space ifp(x, y) = 0 impliesp(x, x) =p(x, y) =p(y, y) = 0 thenx=y.But in a partial metric space ifx=y then p(x, x) = p(x, y) = p(y, y) may not be equal zero. Therefore the partial metric space may not be a b-metric space.
On the other hand, Shukla[21] introduced the notion of a partial b-metric space as follows:
Definition 1.3. [21] LetX be a nonempty set ands≥1 be a given real number. A functionpb:X×X → [0,∞) is called a partialb-metric if for allx, y, z ∈X the following conditions are satisfied:
(i) x=y if and only ifpb(x, x) =pb(x, y) =pb(y, y);
(ii)pb(x, x)≤pb(x, y);
(iii) pb(x, y) =pb(y, x);
(iv)pb(x, y)≤s[pb(x, z) +pb(z, y)]−pb(z, z).
The pair (X, pb) is called a partialb-metric space. The number s≥1 is called the coefficient of (X, pb).
Remark 1.2. The class of partialb-metric space (X, pb) is effectively larger than the class of partial metric space, since a partial metric space is a special case of a partialb-metric space (X, pb) whens= 1. Also, the class of partial b-metric space (X, pb) is effectively larger than the class of b-metric space, since ab-metric space is a special case of a partial b-metric space (X, pb) when the self distancep(x, x) = 0.
The following examples shows that a partial b-metric onX need not be a partial metric, nor a b-metric on X see also [17], [21].
Example1.1. [21] LetX = [0,∞). Define a functionpb :X×X→[0,∞) such thatpb(x, y) = [max{x, y}]2+
|x−y|2 For allx, y∈X. Then (X, pb) is a partialb-metric space on X with the coefficient s= 2>1. But, pb is not ab-metric nor a partial metric onX.
Proposition 1.1. [21] Let X be a nonempty set, and let p be a partial metric and dbe a b-metric with the coefficient s≥1 on X. Then the function pb :X×X →[0,∞) defined by pb(x, y) = p(x, y) +d(x, y) For allx, y∈X, is a partial b-metric on X with the coefficient s.
Proposition 1.2. [21] Let (X, p) be a partial metric space and q ≥ 1. Then (X, pb) is a partial b-metric space with coefficient s= 2q−1, where pb is defined by pb(x, y) = [p(x, y)]q.
Definition 1.4. [14] A function ψ: [0,∞)→ [0,∞) is called an altering distance function if the following properties are satisfied:
1. ψis continuous and nondecreasing;
2. ψ(t) = 0 if and only if t= 0.
On the other hand, Mustafa[17] modify the Definition 1.3 in order that each partialb-metricpb generates ab-metric dpb as follows:
Definition 1.5. [17] LetX be a nonempty set ands≥1 be a given real number. A functionpb:X×X → [0,∞) is a partial b-metric if for allx, y, z∈X the following conditions are satisfied:
(i) x=y if and only ifpb(x, x) =pb(x, y) =pb(y, y);
(ii)pb(x, x)≤pb(x, y);
(iii) pb(x, y) =pb(y, x);
(iv)pb(x, y)≤s[pb(x, z) +pb(z, y)−pb(z, z)] + (1−s2 )(pb(x, x) +pb(y, y)).
The pair (X, pb) is called a partialb-metric space. The number s≥1 is called the coefficient of (X, pb).
Example 1.2 (see also[17]). Let X =R is the set of real numbers. Consider the metric space (X, d) where dis the Euclidean distance metric d(x, y) = |x−y| for allx, y ∈ X. Definepb(x, y) = (x−y)2+ 5 for all x, y∈ X. Thenpb is a partial b-metric on X withs = 2, but it is not a partial metric onX. To see this, Letx= 1,y= 4 and z= 2. Then
pb(1,4) = (1−4)2+ 5 = 14pb(1,2) +pb(2,4)−pb(2,2) = 6 + 9−5 = 10.
Also,pb is not ab-metric since pb(x, x)6= 0 for all x∈X.
Proposition 1.3. [17] Every partial b-metric pb defines a b-metricdpb, where
dpb(x, y) = 2pb(x, y)−pb(x, x)−pb(y, y), for all x, y∈X (1.1) Definition 1.6. [17] A sequence {xn}in a partial b-metric space (X, pb) is said to be:
(i) pb-convergent to a point x∈X if limn→∞pb(x, xn) =pb(x, x);
(ii) a pb-Cauchy sequence if limn,m→∞pb(xn, xm)exists (and is finite);
(iii) A partial b-metric space (X, pb) is said to be pb-complete if every pb-Cauchy sequence {xn} in X pb
converges to a pointx∈X such that
n,m→∞lim pb(xn, xm) = lim
n→∞pb(xn, x) =pb(x, x). (1.2) Lemma 1.1. [17] A sequence{xn} is apb-Cauchy sequence in a partialb−metric space (X, pb) if and only if it is ab-Cauchy sequence in theb-metric space (X, dpb).
Lemma 1.2. [17] A partial b-metric space (X, pb) ispb-complete if and only if theb-metric space (X, dpb) is b-complete. Moreover, limn→∞dpb(xn, xm) = 0 if and only if
n→∞lim pb(x, xn) = lim
n,m→∞pb(x, xm) =pb(x, x). (1.3) Lemma 1.3. [17] Let (X, pb) be a partial b-metric space with the coefficients > 1 and suppose that {xn} and{yn}are convergent to xand y, respectively. Then we have
1
s2pb(x, y)−1
spb(x, x)−pb(y, y) ≤ lim inf
n→∞ pb(xn, yn)
≤ lim sup
n→∞ pb(xn, yn)
≤ spb(x, x) +s2pb(y, y) +s2pb(x, y).
Definition 1.7. [13] Let (X,) be a partially ordered set and T :X → X be a mapping. We say thatT is nondecreasing with respect toif
x, y∈X, xy⇒T xT y.
Definition 1.8. [13] Let (X,) be a partially ordered set. A sequence {xn} is said to be nondecreasing with respect toifxnxn+1 for all n∈N.
Definition 1.9. [17] A triple (X,, pb) is called an ordered partial b-metric space if (X,) is a partially ordered set andpb is a partial b-metric onX.
Definition 1.10. Let (X, pb) be a partialb-metric space andT :X −→Xbe a given mapping. We say that T is α-admissible ifx, y∈X, α(x, y)≥1 implies thatα(T x, T y)≥1. Also we say thatT is Lα-admissible (Rα-admissible) if x, y∈X, α(x, y)≥1 implies that α(T x, y)≥1(α(x, T y)≥1).
Example 1.3. [19] Let X= (0,∞). Define T :X →X and α :X×X →[0,∞) byT x =lnx for allx ∈X and
α(x, y) =
2 if x≥y, 0 if x < y.
Then,T is α-admissible.
Example 1.4. [19] Let X = [0,∞). Define T :X →X and α :X×X → [0,∞) by T x=√
x for allx∈X and
α(x, y) =
ex−y if x≥y, 0 if x < y.
Then,T is α-admissible.
2. Main result
We now introduce theα-ψ-ϕ-contractive self mapping on partialb-metric space.
Definition 2.1. Let (X, pb) be a partial b-metric space with the coefficient s≥1. We say that a mapping T : X → X is an α-ψ-ϕ-contractive mapping if there exist two altering distance functions ψ, ϕ and α:X×X:→[0,∞) such that
α(x, y)ψ(spb(T x, T y))≤ψ(MsT(x, y))−ϕ(MsT(x, y)) (2.1) for all comparablex, y∈X.
where
MsT(x, y) =max{pb(x, y), pb(x, T x), pb(y, T y),pb(x, T y) +pb(y, T x)
2s }. (2.2)
Theorem 2.1. Let (X,, pb) be a pb-complete ordered partial b-metric space with the coefficients≥1. Let T :X →X be an an α-ψ-ϕ-contractive mapping. Suppose that the following conditions hold:
(1)T is α-admissible and Lα-admissible (orRα-admissible );
(2) there existsx1∈X such thatx1T x1 and α(x1, T x1)≥1;
(3)T is continuous, nondecreasing, with respect to and if Tnx1 →z thenα(z, z)≥1 . Then,T has a fixed point.
Proof. Let x1 ∈X such that x1 T x1 and α(x1, T x1) ≥1. Define a sequence {xn} in X by xn+1 =T xn for all n≥ 1. We have x2 =T x1 T x2 = x3 since x1 T x1 and T is nondecreasing. Also, x3 = T x2 T x3 =x4 sincex2 T x2 and T is nondecreasing. By induction, we get
x1x2x3 · · · xnxn+1 · · ·.
Ifxn=xn+1 for somen∈N, thenx=xnis a fixed point of T and the proof is finished. So we may assume thatxn6=xn+1 for all n∈N. SinceT isα-admissible, we deduce
α(x1, T x1) =α(x1, x2)≥1⇒α(T x1, T x2) =α(x2, x3)≥1.
By induction onnwe get
α(xn, xn+1)≥1 (2.3)
for all n∈N.
Hence, by applying the α-ψ-ϕ-contractive condition and using (2.3) for all n∈Nwe get ψ(spb(xn+1, xn+2)) ≤ α(xn, xn+1)ψ(spb(T xn, T xn+1))
≤ ψ(MsT(xn, xn+1))−ϕ(MsT(xn, xn+1)) (2.4) where
MsT(xn, xn+1) = max{pb(xn, xn+1), pb(xn, T xn), pb(xn+1, T xx+1), pb(xn, T xn+1) +pb(xn+1, T xn)
2s }
= max{pb(xn, xn+1), pb(xn+1, xx+2), pb(xn, xn+2) +pb(xn+1, xn+1)
2s }
≤ max{pb(xn, xn+1), pb(xn+1, xx+2),
spb(xn, xn+1) +spb(xn+1, xn+2) + (1−s)pb(xn+1, xn+1)
2s }
= max{pb(xn, xn+1), pb(xn+1, xx+2)} (2.5)
From (2.4)and (2.5) we get
ψ(pb(xn+1, xn+2)) ≤ ψ(max{pb(xn, xn+1), pb(xn+1, xx+2)})
−ϕ(max{pb(xn, xn+1), pb(xn+1, xx+2)}) (2.6) Assume that
max{pb(xn, xn+1), pb(xn+1, xx+2)}=pb(xn+1, xx+2) then by using properties ofϕ, we deduce
ψ(pb(xn+1, xn+2))≤ψ(pb(xn+1, xn+2))−ϕ(pb(xn+1, xn+2))
< ψ(pb(xn+1, xn+2)), which is a contradiction. Thus,
ψ(pb(xn+1, xn+2)) ≤ ψ(pb(xn, xn+1))−ϕ(pb(xn, xn+1)). (2.7) So the sequence{pb(xn+1, xn+2)}is nonnegative and nondecreasing for all n∈N. Hence there existsr≥0 such that
n→∞lim pb(xn+1, xn+2) =r.
Lettingn→ ∞in (2.7) , we have
ψ(r)≤ψ(r)−ϕ(r)≤ψ(r).
Therefore, ϕ(r) = 0, and hence r= 0. Thus,
n→∞lim pb(xn+1, xn+2) = 0. (2.8)
Now, we show that{xn}is a Cauchy sequence in (X, pb) which is equivalent to show that{xn}is ab-Cauchy sequence in (X, dpb). Assume not, that is, {xn} is not a b-Cauchy sequence in (X, dpb). Then there exist ε >0 such that, fork >0, there existn(k)> m(k)> kfor which we can which we can find two subsequences {xn(k)}and {xm(k)} of {xn}such that n(k) is the smallest index for which
dpb(xm(k), xn(k))≥ε, (2.9)
and
dpb(xm(k), xn(k)−1)< ε. (2.10)
Then we have
ε≤dpb(xm(k), xn(k)) ≤ sdpb(xm(k), xn(k)−1) +sdpb(xn(k)−1, xn(k))
< sε+sdpb(xn(k)−1, xn(k)). (2.11)
Taking the upper limit for (2.10) ask→ ∞, we have ε
s ≤lim inf
k→∞ dpb(xm(k), xn(k)−1)≤lim sup
k→∞
dpb(xm(k), xn(k)−1)≤ε. (2.12)
Also, from (2.11)and (2.12), we obtain
ε≤lim sup
k→∞
dpb(xm(k), xn(k))≤sε.
By using the triangular inequality and we deduce,
dpb(xm(k)+1, xn(k))≤sdpb(xm(k)+1, xm(k)) +sdpb(xm(k), xn(k))
≤ sdpb(xm(k)+1, xm(k)) +s2dpb(xm(k), xn(k)−1) +s2dpb(xn(k)−1, xn(k))
≤ sdpb(xm(k)+1, xm(k)) +s2ε+s2dpb(xn(k)−1, xn(k)), by taking the upper limit ask→ ∞in the above inequality, we get
lim sup
k→∞
dpb(xm(k)+1, xn(k))≤s2ε.
Finally,
dpb(xm(k)+1, xn(k)−1)≤sdpb(xm(k)+1, xm(k)) +sdpb(xm(k), xn(k)−1)
≤ sdpb(xm(k)+1, xm(k)) +sε.
Also, by taking the upper limit as k→ ∞ in the above inequality, we get lim sup
k→∞
dpb(xm(k)+1, xn(k)−1)≤sε.
By using the definition ofdpb and (2.8), we get lim sup
k→∞
dpb(xm(k), xn(k)−1) = 2 lim sup
k→∞
pb(xm(k), xn(k)−1).
Hence, ε
2s ≤lim inf
k→∞ pb(xm(k), xn(k)−1)≤lim sup
k→∞
pb(xm(k), xn(k)−1)≤ ε
2, (2.13)
Similarly, lim sup
k→∞
pb(xm(k), xn(k))≤ sε
2, (2.14)
ε
2s ≤lim sup
k→∞
pb(xm(k)+1, xn(k)), (2.15)
lim sup
k→∞
pb(xm(k)+1, xn(k)−1)≤ sε
2 . (2.16)
Since T isLα-admissible and using (2.3), we obtainα(xm(k), xn(k)−1)≥1.
By using (2.1) we get
ψ(spb(xm(k)+1, xn(k)))≤α(xm(k), xn(k)−1)ψ(spb(T xm(k), T xn(k)−1))
≤ψ(MsT(xm(k), xn(k)−1))−ϕ(MsT(xm(k), xn(k)−1)), (2.17) where
MsT(xm(k), xn(k)−1) = max{pb(xm(k), xn(k)−1), pb(xm(k), T xm(k)), pb(xn(k)−1, T xn(k)−1), pb(xm(k), T xn(k)−1) +pb(xn(k)−1, T xm(k))
2s }
= max{pb(xm(k), xn(k)−1), pb(xm(k), xm(k)+1), pb(xn(k)−1, xn(k)), pb(xm(k), xn(k)) +pb(xn(k)−1, xm(k)+1)
2s } (2.18)
Taking the upper limit as k→ ∞ in the above inequality using (2.8),(2.13),(2.14)and (2.16) we get lim sup
k→∞
MsT(xm(k), xn(k)−1) = max{lim sup
k→∞
pb(xm(k), xn(k)−1),lim sup
k→∞
pb(xm(k), xm(k)+1), lim sup
k→∞
pb(xn(k)−1, xn(k)),
lim supk→∞pb(xm(k), xn(k)) + lim supk→∞pb(xn(k)−1, xm(k)+1)
2s }
= max{lim sup
k→∞
pb(xm(k), xn(k)−1),0,0,
lim supk→∞pb(xm(k), xn(k)) + lim supk→∞pb(xn(k)−1, xm(k)+1)
2s }
≤ max{ε 2,ε
2}
= ε
2. (2.19)
Next, by taking the upper limit in (2.17) ask→ ∞and using(2.15) and (2.19) we obtain ψ(sε
2s)≤ψ(lim sup
k→∞
spb(xm(k)+1, xn(k)))
≤ψ(lim sup
k→∞
MsT(xm(k), xn(k)−1))−lim inf
k→∞ ϕ(MsT(xm(k), xn(k)−1)),
≤ψ(ε
2)−ϕ(lim inf
k→∞ MsT(xm(k), xn(k)−1)), which implies that
ϕ(lim inf
k→∞ MsT(xm(k), xn(k)−1)) = 0, so
lim inf
k→∞ MsT(xm(k), xn(k)−1)) = 0,
and by using (2.17) we obtain,
lim inf
k→∞ pb(xm(k), xn(k)−1) = 0.
Therefore,
lim inf
k→∞ dpb(xm(k), xn(k)−1) = 0,
which is a contradiction with (2.13). Thus, the sequence is ab-Cauchy in theb-metric space (X, dpb). Since (X, pb) is pb-complete, then (X, dpb) is a b-complete b-metric space. So, it follows from the completeness that there existz∈X such that,
n→∞lim dpb(xn, z) = 0.
Therefore, by using (2.8), the condition pb(xn, xn)≤pb(z, xn) and limn→∞pb(xn, xn) = 0 we get
n→∞lim pb(xn, z) = lim
n→∞pb(xn, xn) =pb(z, z) = 0.
By using the triangular inequality, we obtain
pb(z, T z)≤spb(z, T xn) +spb(T xn, T z).
So by taking limit asn→ ∞ in the above inequality and using the continuity ofT we get pb(z, T z)≤s lim
n→∞pb(z, xn+1) +s lim
n→∞pb(T xn, T z) =spb(T z, T z). (2.20) Since α(z, z)≥1 and using (2.1) we get
ψ(spb(T z, T z))≤α(z, z)ψ(spb(T z, T z))≤ψ(MsT(z, z))−ϕ(MsT(z, z)).
where
MsT(z, z) =max{pb(z, z), pb(z, T z), pb(z, T z),pb(z, T z) +pb(z, T z)
2s }=pb(z, T z).
So
ψ(spb(T z, T z))≤α(z, z)ψ(spb(T z, T z))≤ψ(pb(z, T z))−ϕ(pb(z, T z)). (2.21) Sinceψis nondecreasingspb(T z, T z)≤pb(z, T z) andspb(T z, T z) =pb(z, T z),we deduceϕ(pb(z, T z)) = 0., Hence we have pb(T z, z) = pb(T z, T z) = pb(z, z) = 0 and T z = z. Thus, z is a fixed point of T. This completes the proof.
In our next theorem we omit the condition of continuity in Theorem 2.1.
Theorem 2.2. Let (X,, pb) be a pb-complete ordered partial b-metric space with the coefficients≥1. Let T :X →X be an an α-ψ-ϕ-contractive mapping. Suppose that the following conditions hold:
(1)T is α-admissible and Lα-admissible (orRα-admissible );
(2) there existsx1∈X such thatx1T x1 and α(x1, T x1)≥1;
(3)T is nondecreasing, with respect to ;
(4) If{xn} is a sequence inX such thatxnxfor all n∈N, α(xn, xn+1)≥1 andxn→x∈X, asn→ ∞, thenα(xn, x)≥1 for alln∈N;
Then,T has a fixed point.
Proof. Following the proof of Theorem 2.1, we know that the sequence {xn}defined by xn+1=T xn for all n ∈ N,is an increasing pb-Cauchy sequence in the pb-complete b-metric space (X, pb).It follows from the completeness of (X, pb) that there exists z∈X such that limn→∞xn=z. Using the assumption onX, we
deduce xn z for alln∈N.So it is enough to show f z =z. Now, by using (2.1) and α(xn, x) ≥1 for all n∈N,we have
ψ(spb(xn+1, T z))≤α(xn, z)ψ(spb(T xn, T z))
≤ψ(MsT(xn, z))−ϕ(Msf(xn, z)), (2.22) where
MsT(xn, z) =max{pb(xn, z), pb(xn, T xn), pb(z, T z),pb(xn,T z)+p2sb(T xn,z)}
≤max{pb(xn, z), pb(xn, xn+1), pb(z, T z),pb(xn,T z)+p2sb(xn+1,z)}. (2.23) By taking the limit asn→ ∞in above inequality and using Lemma 1.3, we get
pb(z, T z)
2s2 = min{pb(z, T z),
pb(z,T z) s
2s }
≤ lim inf
n→∞ MsT(xn, z)
≤ lim sup
n→∞ MsT(xn, z)
≤ max{pb(z, T z),spb(z, T z)
2s }=pb(z, T z). (2.24)
Again, by using (2.22) and taking the upper limit asn→ ∞
ψ(spb(xn+1, T z))≤α(xn, z)ψ(spb(T xn, T z))
≤ψ(Msf(xn, z))−ϕ(Msf(xn, z)), and using Lemma 1.3, we get
ψ(pb(z, T z)) =ψ(s1
spb(xn+1, T z))
≤ψ(slim sup
n→∞
pb(xn+1, T z))
≤ψ(lim sup
n→∞ Msf(xn, z))−lim inf
n→∞ ϕ(MsT(xn, z))
≤ψ(pb(z, T z))−ϕ(lim inf
n→∞ MsT(xn, z)).
Therefore, ϕ(lim infn→∞MsT(xn, z)) ≤ 0, it means that lim infn→∞MsT(xn, z)) = 0. Thus, from (2.24)we getz=T z, and hence zis a fixed point ofT. This completes the proof.
Example 2.1. LetX= [1,∞) be equipped with the partial orderdefined by xy⇐⇒x≤y
and with the functional pb :X×X → [0,∞) defined by pb(x, y) = |x−y|2+ 2 For all x, y ∈ X. Clearly, (X, pb) is a partial complete b-metric space withs= 2. Define the mapping T :X→X by
T x= (x+6
4 if 1≤x≤2,
x2
2 ifx >2, and α:X×X→[0,∞) by
α(x, y) =
1 if x, y∈[1,2], 0 otherwise.
and taking the altering distance functions ψ(t) =t and ϕ(t) =
( (t−1)2
2 if x, y∈[1,2],
t
4 t >2.
ThenT is continuous and increasing, 1T1.For Checking the contraction condition (2.1) for all comparable x, y∈X. Letx= 1 and y= 4, we get
ψ(2pb(T1, T4)) =ψ(2pb(7
4,8)) = 641 8 108
8
=ψ(18)−ϕ(18) = 18− 18
4 =ψ(MsT(1,4))−ϕ(MsT(1,4)).
We will prove the following:
i) T :X→X is anα-ψ-ϕ-contractive mapping, with ψ(t) =t for allt≥0;
ii) T is α−admissible;
iii) there exists x1 = 1∈X and x1 T x1, such thatα(x1, T x1)≥1;
iv) If {xn}∞n=1 is a sequence in X such that α(xn, xn+1) ≥1 and xn→ x asn→ ∞,then α(xn, x) ≥1 for all n∈N;
Proof. i) ClearlyT isα-ψ-ϕ-contractive mapping withψ(t) =t for allt≥0,since for all x, y∈X, α(x, y)ψ(spb(T x, T y)) =ψ(2pb(x+ 6
4 ,y+ 6
4 )) = 2|x+ 6
4 −y+ 6
4 |2+ 2 = 1
8|x−y|2+ 2, while without loss of generality if 1≤y≤x≤2,then
α(x, y)ψ(spb(T x, T y)) = ψ(2pb(x+ 6 4 ,y+ 6
4 ))
= 2|x+ 6
4 −y+ 6
4 |2+ 2 = 1
8|x−y|2+ 2
≤ 9
4 = 3−3
4 =ψ(3)−ϕ(3)
= ψ(MsT(x, y))−ϕ(MsT(x, y)).
ii) Let (x, y) ∈X×X such that α(x, y) ≥1. From the definition ofT and α we have bothT x = x+64 ,and T y= y+64 are in [1,2], so we have α(T x, T y) = 1≥1. Then T is anα-admissible.
iii) Takingx1 = 1∈X,we have
α(x1, T x1) =α(1, T1) =α(1,7
4) = 1≥1.
iv) let {xn}be a sequence inX such thatα(xn, xn+1)≥1 for alln∈N andxn→x∈X asn→ ∞. Since α(xn, xn+1)≥1 for alln∈Nand by the definition ofα,we havexn∈[1,2] for alln∈Nandx∈[1,2].Then α(xn, x) = 1≥1. Now, all the hypothesis of Theorem 2.1 are satisfied. Therefore, T has a fixed point.
Example 2.2. LetX= [0,∞) be equipped with the partial orderdefined by xy⇐⇒x≤y
and with the functional pb :X×X →[0,∞) defined by pb(x, y) = [max{x, y}]2 For all x, y∈ X. Clearly, (X, pb) is a partial ordered completeb-metric space withs= 2. Define the mapping T :X →X by
T x= ( x
√ 2√
1+x if 0≤x≤1,
x
2 ifx >1,
and α:X×X→[0,∞) by
α(x, y) =
1 if x, y∈[0,1], 0 otherwise.
and taking the altering distance functions ψ(t) =t and ϕ(t) =
( t√ t 1+√
t if t∈[0,1],
t
2 t >1.
ThenT is continuous and increasing, 0T0.For Checking the contraction condition (2.1) for all comparable x, y∈X. Letx= 0 and y= 4, we get
ψ(2pb(T0, T2)) =ψ(2pb((0,2)) = 82 = 4−2
=ψ(4)−ϕ(4) =ψ(MsT(0,2))−ϕ(MsT(0,2)).
We will prove the following:
i) T :X→X is anα-ψ-ϕ-contractive mapping, with ψ(t) =t for allt≥0;
ii) T is α−admissible;
iii) there exists x1 = 0∈X and x1 T x1, such thatα(x1, T x1)≥1;
iv) If {xn}∞n=1 is a sequence in X such that α(xn, xn+1) ≥1 and xn→ x asn→ ∞,then α(xn, x) ≥1 for all n∈N;
Proof. i) ClearlyT isα-ψ-ϕ-contractive mapping withψ(t) =t for allt≥0,since for all x, y∈X, α(x, y)ψ(spb(T x, T y))≤ψ(MsT(x, y))−ϕ(MsT(x, y))
Since,
α(x, y)ψ(spb(T x, T y)) = ψ(2pb( x
√ 2√
1 +x, y
√ 2√
1 +y))
= 2[max{ x
√ 2√
1 +x, y
√ 2√
1 +y}]2
= x2
(1 +x) and
MsT(x, y) = max{x2, x2, y2,
x2+ [max{y,√ x
2√ 1+x}]2
4 }=x2
Thus
α(x, y)ψ(spb(T x, T y)) = x2
(1 +x) ≤x2− x3
1 +x =ψ(x2)−ϕ(x2) =ψ(MsT(x, y))−ϕ(MsT(x, y)).
ii) Let (x, y)∈X×Xsuch thatα(x, y)≥1. From the definition ofT andαwe have bothT x= √ x
2√
1+x,and T y= √ y
2√
1+y are in [0,1], so we have α(T x, T y) = 1≥1. ThenT is anα-admissible.
iii) Takingx1 = 0∈X,we have
α(x1, T x1) =α(0, T0) =α(0,0) = 1≥1.
iv) let {xn}be a sequence inX such thatα(xn, xn+1)≥1 for alln∈N andxn→x∈X asn→ ∞. Since α(xn, xn+1)≥1 for alln∈Nand by the definition ofα,we havexn∈[0,1] for alln∈Nandx∈[0,1].Then α(xn, x) = 1≥1.
Now, all the hypothesis of Theorem 2.1 are satisfied. Therefore, T has a fixed point
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